The statement for, "endothermic reaction has high energy for breaking bonds" has been true.
The chemical reaction has been differentiated as the endothermic and exothermic reaction based on the energy absorbed or release by the system.
The chemical compounds have been consisted of the conserved energy that has been released by the chemical when they break the bond. The energy has been absorbed by the bonds for the formation of new bonds.
Do endothermic energy has high energy released?The endothermic reaction has been given as the reaction that has been absorbed the energy from the surrounding for the breaking of bonds.
In the endothermic reaction, the energy for breaking bonds has been higher than the energy required to form new bonds in the products.
Thus, the system absorbs the energy from the surrounding and result in the rise in temperature of the system.
Thus, the statement for, "endothermic reaction has high energy for breaking bonds" has been true.
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https://brainly.com/question/2192784
Answer:
True
Explanation:
Got it right on the quiz
How to apply Raoult's law to real solutions Consider mixing a liquid with a vapor pressure of 100 torr with an equimolar amount of a liquid with a vapor pressure of 200 torr. The resulting solution would be predicted to have a vapor pressure of 150 torr if it behaved ideally. If, however, the interactions between the different components are not similar we can see positive or negative deviations from the calculated vapor pressure. An actual vapor pressure greater than that predicted by Raoult's law is said to be a positive deviation and an actual vapor pressure lower than that predicted by Raoult's law is a negative deviation. Part A Imagine a solution of two liquids in which the molecules interact less favorably than they do in the individual liquids. Will this solution deviate positively from, deviate negatively from, or ideally follow Raoult's law?'
Answer:
[tex]\boxed{\text{positive deviation}}[/tex]
Explanation:
[tex]\text{If the molecules interact less favourably in the solution than in the individual}\\\text{liquids, they can break away more easily than they can in the separate liquids.}\\\text{There will be more molecules in the vapour phase than you would expect.}\\\text{The solution will show a } \boxed{\textbf{positive deviation}} \text{ from Raoult's Law.}[/tex]
In a solution of pure water, the dissociation of water can be expressed by the following: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) The equilibrium constant for the ionization of water, Kw, is called the ion-product of water. In pure water at 25 °C, Kw has a value of 1.0 × 10−14. The dissociation of water gives one H3O+ ion and one OH− ion and thus their concentrations are equal. The concentration of each is 1.0 × 10−7 M. Kw = [H3O+][OH−] Kw = (1.0 × 10-7)(1.0 × 10-7) = 1.0 × 10-14 [H3O+][OH−] = 1.0 × 10-14 A solution has a [OH−] = 3.4 × 10−5 M at 25 °C. What is the [H3O+] of the solution? ANSWER 2.9 × 10−9 M 2.9 × 10−15 M 3.4 × 109 M 2.9 × 10−10 M I DON'T KNOW YET
Answer:
[H₃O⁺] = 2.90 × 10⁻¹⁰ MExplanation:
1) Ionization equilibrium equation: given
H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)2) Ionization equilibrium constant, at 25°C, Kw: given
Kw = 1.0 × 10⁻¹⁴3) Stoichiometric mole ratio:
As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:
[H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C and you need to calculate what the [H₃O⁺(aq)] is.
Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:
Kw = 1.0 × 10⁻¹⁴ M², and Kw = [H3O⁺] [OH⁻]Then you can substitute the known values and solve for the unknown:
1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M ⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ MAs you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.
Answer:
1. pH = -log[H^+]
2. Acidic: pH < 7.00, neutral: pH = 7.00, basic: pH > 7.00
3. pH + pOH = 14.00 at room temperature
Explanation:
In the reversible reaction: 2NO2 (g) ⇌ N2O4 (g) the formation of dinitrogen tetroxide releases heat and the formation of nitrogen dioxide absorbs heat. If the reaction is at equilibrium and the temperature increases, what will the effect be?
A. Nitrogen dioxide absorbs heat and changes from gas to liquid.
B. The equilibrium will shift so that there is more nitrogen dioxide.
C. The equilibrium will shift so that there is more dinitrogen tetroxide.
D. Dinitrogen tetroxide absorbs heat and changes from gas to liquid.
Answer: Option (B) is the correct answer.
Explanation:
According to Le Chatelier's principle, any disturbance causes in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.
For example, [tex]2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g)[/tex]
When we increase the temperature then the reaction will shift in a direction where there will be decrease in temperature.
This, means that the reaction will shift in the backward direction.
Thus, we can conclude that if the reaction is at equilibrium and the temperature increases, the equilibrium will shift so that there is more nitrogen dioxide.
Answer:
B. The equilibrium will shift so that there is more nitrogen dioxide.
Explanation:
Hello,
By considering that the mentioned reaction is exothermic, it means that the heat is like a product, if heat is added (increase the temperature) the equilibrium will shift leftwards, contributing to the inverse reaction; it is the formation of more nitrogen dioxide by recalling the Le Chatelier's principle.
Best regards.
Hydrogen sulfide gas (H2S) is a highly toxic gas that is responsible for the smell of rotten eggs. The volume of a container of hydrogen sulfide is 44.2mL. After the addition of more hydrogen sulfide, the volume increases to 98.5mL under constant pressure and temperature. The container now holds 1.97×10−3mol of the gas. How many grams of hydrogen sulfide were in the container initially? Give your answer in three significant figures.
Answer:
[tex]\boxed{\text{0.0301 g}}[/tex]
Step-by-step explanation:
1. Calculate the initial moles of H₂S
We can use Avogadro's law: the number of moles of a gas is directly proportional to the volume if the pressure and temperature are constant.
[tex]\dfrac{n_{1} }{V_{1}} = \dfrac{n_{2}}{V_{2}}\\\\\dfrac{n_{1}}{\text{44.2 mL}} = \dfrac{1.97 \times 10^{-3} \text{ mol}}{\text{98.5 mL }}\\\\\dfrac{n_{1}}{44.2} = 2.00 \times 10^{-5} \text{ mol}\\\\n_{1} = 44.2 \times 2.00 \times 10^{-5} \text{ mol} = 8.84 \times 10^{-4} \text{ mol}[/tex]
2. Calculate the initial mass of H₂S
[tex]\text{Mass of H$_{2}$S} = 8.84 \times 10^{-4}\text{ mol H$_{2}$S} \times \dfrac{\text{34.08 g H$_{2}$S}}{\text{1 mol H$_{2}$S}} = \text{0.0301 g H$_{2}$S}\\\\\text{The container initially held }\boxed{\textbf{0.0301 g H$_{2}$S}}[/tex]
The reaction 2HI → H2 + I2 is second order in [HI] and second order overall. The rate constant of the reaction at 700°C is 1.57 × 10−5 M −1s−1. Suppose you have a sample in which the concentration of HI is 0.75 M. What was the concentration of HI 8 hours earlier?
Answer:
1.135 M.
Explanation:
For the reaction: 2HI → H₂ + I₂,The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².
To solve this problem, we can use the integral law of second-order reactions:1/[A] = kt + 1/[A₀],
where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),
t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),
[A₀] is the initial concentration of HI ([A₀] = ?? M).
[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).
∵ 1/[A] = kt + 1/[A₀],
∴ 1/[A₀] = 1/[A] - kt
∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.
∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.
So, the concentration of HI 8 hours earlier = 1.135 M.
Final answer:
To find the concentration of HI 8 hours earlier in a second-order reaction, we utilize the integrated rate law specific for second-order reactions and solve for the initial concentration using the given rate constant and the time in seconds.
Explanation:
The reaction 2HI → H2 + I2 is second order in [HI] and second order overall, meaning the rate equation can be written as Rate = k[HI]2. Given the rate constant (k) of 1.57 × 10−5 M−1s−1 at 700°C and the current concentration of 0.75 M, we need to calculate the concentration 8 hours earlier. To solve this, we can use the integrated rate law for a second-order reaction:
[HI]−1 - [HI]0−1 = kt
Where [HI]0 is the initial concentration, [HI] is the concentration after time t, k is the rate constant, and t is the time in seconds. We have:
[HI] = 0.75 M
k = 1.57 × 10−5 M−1s−1
t = 8 hours × 3600 seconds/hour = 28800 seconds
Plugging these values into the integrated rate equation, we calculate [HI]0. Finally, we can find the concentration 8 hours earlier.
Ions are formed when atoms a) gain or lose protonsb) gain or lose electronsc) gain or lose neutrons d) each of these results in ion formatione) none of these results in ion formation
Answer:
b) Gain or lose electrons
Explanation:
An ion is an electrically charged particle. For an atom to be charged, it must have gained or lost electron in the process and therefore, it becomes an ion.
The loss or gain of electrons is what makes an atom charged and eventually becomes an ion.
A positively charged ion is one that has lost an electron and it is called a cation. In such an ion, the number of electrons are lesser than those of protons. This is why they are cations
A negatively charged ion is one that has gained electrons. They are called anions. In such an ion, the number of electrons are greater than that of protons.
You need to know the volume of water in a small swimming pool, but, owing to the pool's irregular shape, it is not a simple matter to determine its dimensions and calculate the volume. To solve the problem, you stir in a solution of a dye (1.00 g of methylene blue, C16H18ClN3S, in 50.0 mL of water). After the dye has mixed with the water in the pool, you take a sample of the water. Using a spectrophotometer, you determine that the concentration of the dye in the pool is 3.74 × 10-8 M. What is the volume of water in the pool in liters? The molar mass of methylene blue is 319.854 g/mol.
Answer:
83,700 liter (rounded to 3 significant figures)Explanation:
1) Data:
a) Solution of a dye:
m₁ = 1.00 g of methylene blue, C₁₆H₁₈ClN₃SV₁ = 50.0 mL MM₁ = 319.854 g/molb) Water of the pool, after mixing the dye:
M₂ = 3.74 × 10⁻⁸ M. V₂ = ?2) Formulae:
a) Molarity: M = n / V in liter
b) Molar mass: MM = mass in grams / number of moles
3) Solution:
a) Number of moles of methylene blue in the prepared solution:
n₁ = mass in grams / molar mass = = 1.00 g / 319.854 g/mol = 0.00313 molb) Number of moles of methylen blue in the pool = number of moles of methylen blue in the prepared solution
n₂ = n₁ = 0.00313 molc) Volume of the pool:
M₂ = n₂ / V₂ ⇒ V₂ = n₂ / M₂ = 0.00313 mol /( 3.74 × 10⁻⁸ M) = 83,700 literAnd that is the answer, which has to be rounded to 3 significant figures, since the data are expressed with 3 significant figures.
Note: For all the effects, the 50.0 mL of the dye solution are neglictible in front of the volume of the pool.
Final answer:
To calculate the volume of water in an irregularly shaped pool using a known concentration of methylene blue dye, the dilution formula is applied with the dye's initial concentration and mass, resulting in the pool's volume being 83,689 liters.
Explanation:
To find the volume of water in the pool using the concentration of methylene blue dye, first, determine the amount of dye initially used to disperse in the pool water. Since 1.00 g of methylene blue was dissolved in 50.0 mL of water, and knowing the molar mass of methylene blue is 319.854 g/mol, we can calculate the molarity of the initial solution before it was diluted in the pool.
Moles of methylene blue = mass (g) / molar mass (g/mol) = 1.00 g / 319.854 g/mol = 0.00313 moles.
Thus, the initial concentration of the dye in the solution dispersed in the pool is 0.00313 moles / 0.050 L = 0.0626 M.
When this dye disperses evenly throughout the pool, its concentration decreases to 3.74 × 10-8 M. The total volume of the pool water can be calculated by applying the concept of dilution, where the moles of dye remain constant but the volume changes:
Initial moles = final molesInitial concentration × initial volume = Final concentration × final volume0.0626 M × 0.050 L = 3.74 × 10-8 M × Final volumeFinal volume = (0.0626 M × 0.050 L) / (3.74 × 10-8 M)Final volume = 83,689 LTherefore, the volume of water in the pool is 83,689 liters.
The half-lives of unstable isotopes vary from milliseconds to billions of years
True or false
Answer:
True
Explanation:
Answer:
True
Explanation:
The half lives of unstable isotopes of elements vary from miliseconds to billions of years, for example Bismuth-209 has the longest half life known by the human and the scientists and it has still billions of years to this day, and there are elements that cannot withstand seconds before decaying into another element, so the statement is True.
What coefficients must be added to balance the following reaction?
_____ Pb + _____ H3PO4 yields _____ H2 + _____ Pb3(PO4)2
A: 3, 2, 1, 1
B: 3, 2, 2, 1
C: 3, 1, 3, 1
D: 3, 2, 3, 1
D: 3, 2, 3, 1 The Pb3(PO4) comes from having 3 Pb and 2 H3PO4. The H2 comes from the 2H3PO4
Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of their practical exam is to determine this integer x. They are working in pairs, though Hermione is doing most of the work. This should be discouraged! After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4. What is the unknown integer x. Round the answer to the nearest integer.
Answer:
The unknown integer X in the formula is 5.Explanation:
1) Data:
a) Mass of CuSO₄ ∙ XH₂O = 1.50 g
b) Mass of CuSO₄ = 0.96
c) X = ?
2) Additional needed data:
a) Molar Mass of CuSO₄ = 159,609 g/mol
b) Molar mass of H₂O = 18,01528 g/mol
3) Chemical principles and formulae used:
a) Law of conservation of mass
b) Molar mass = mass in grams / number of moles = m / n
4) Solution:
a) Law of conservation of mass:
Mass of CuSO₄ ∙ XH₂O = mass of CuSO₄ + mass of H₂O1.50g = 0.96g + mass of H₂O ⇒ mass of H₂O = 1.50g - 0.96g = 0.54gb) Moles
n = m / molar massCuSO₄: n = 0.96g / 159.609 g/mol = 0.0060 molH₂O: 0.54g / 18.01528 g/mol = 0.030 molc) Proportion:
Divide both mole amounts by the least of the two numbers, i.e. 0.0060
CuSO₄: 0.0060 / 0.0060 = 1H₂O: 0.030 mol / 0.0060 = 5Then, the ratio of CuSO₄ to H₂O is 1 : 5 and the chemical formula is:
CuSO₄ . 5H₂O.Hence, the value of X is 5.
Answer:
CuSO4*5H2O
X = 5
Explanation:
Step 1: Data given
Mass of the copper(II)sulfate hydrate = 1.50 grams
After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4.
Step 2: Calculate mass of water
Mass of water = mass of hydrate - mass of anhydrate
Mass water = 1.50 grams - 0.96 grams
Mass water = 0.54 grams
Step 3: Calculate moles of water
Moles H2O = mass / molar mass
Moles H2O = 0.54 grams / 18.02 g/mol
Moles H2O = 0.030 moles
Step 4: Calculate moles of CuSO4
Moles CuSO4 = 0.96 grams / 159.61 g/mol
Moles CuSO4 = 0.0060 moles
Step 5: Calculate mol ratio
We divide by the smallest amount of moles
H2O : 0.030 / 0.060 = 5
CuSO4: 0.0060 / 0.0060 = 1
For 1 mol CuSO4 we have 5 moles of H2O
CuSO4*5H2O
X = 5
Electrons are ejected from a surface with speeds ranging up to 1380 km/s when light with a wavelength of 173 nm is used. What is the work function of this surface? The speed of light is 2.99792 × 108 m/s and Planck’s constant is 6.62607 × 10−34 J · s. Answer in units of eV.
Answer:
ΔE(work)= 692 Kj/mole e⁻ = 4.32 x 10²⁴ eV/mole e⁻
Explanation:
ΔE = hc/λ = (6.63 x 10⁻³⁴j·s)(2.99792 x 10⁸m/s)/(1.73 x 10⁻⁷m) = 1.15 x 10⁻¹⁸j/e⁻ x 6.023 x 10²³ e⁻/mole = 691,999 j/mole e⁻ = 692 Kj/mole e⁻ x 6.24 x 10²¹ eV/KJ = 4.32 x 10²⁴ eV/mole e⁻
What does it mean when a mineral has a definite chemical composition?
Answer:
For a substance to classify as a mineral, it must lie within certain parameters. It should be an inorganic solid, that is naturally occurring in nature (not synthesized), with an ordered internal structure and a definite chemical composition.
By definite chemical composition, geologists mean that the mineral must be have chemical constituents that have an unvarying chemical composition, or a chemical composition that oscillates withing a very limited and specific range.
An example is the mineral, halite. It has a chemical composition of one sodium atom and one chloride atom, represented as NaCl and is unchanging in this composition throughout nature.
Hope this helpsA) The breaking of chemical bonds
B) physical changes in Atoms
C) Chemical reactions
D) The nuclei of unstable isotopes
Question, where does radioactivity come from?
Answer:
D) The nuclei of unstable isotopes
Explanation:
For example, the carbon-14 in carbon dating is radioactive.
When it decays, a neutron in the nucleus is converted into a proton, an electron, and an antineutrino.
[tex]_{0}^{1}\text{n} \longrightarrow \, _{1}^{1}\text{p} + \, _{-1}^{0}\text{e} + \overline{\nu}_{\text{e}}[/tex]
How are energy and work related?
O
A. Work and energy are the same.
B. Energy is the capacity to do work.
C. Energy is the force needed to do work.
D. Work times energy is force.
Answer:
B
Explanation:
When work is done, energy is transferred between systems, or transformed from one type of energy into another type. Energy shares the same units of measure as work.
Work and energy are directly related in physics - the amount of work done is the same as the energy transferred. Energy can be defined as the capacity to do work.
Explanation:Work and energy are directly related concepts in physics. The amount of work done is equivalent to the amount of energy transferred. This is because, by definition, work is done when a force acts on an object to move it some distance, and this process requires energy. Therefore, energy can be viewed as the capacity to do work. If an entity does not contain energy, it cannot perform work. This concept is fundamental to understanding the physics of energy transfer and conservation.
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Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Standard reduction potentials (E°red) may be found here.'1) Pt(s)+Fe2+(aq)\rightleftharpoonsPt2+ (aq)+Fe(s)[Fe2+]=0.0066M [Pt2+]=0.057ME= ? V2) Cu(s)+2Ag+(aq)\rightleftharpoonsCu2+(aq)+2Ag(s)[Cu2+]=0.013M [Ag+]=0.013ME= ?V3) Co2+(aq)+Ti3+(aq)\rightleftharpoonsCo3+(aq)+Ti2+(aq)[Co2+]=0.050M [Co3+]=0.030M[Ti3+]=0.0055M [Ti2+]=0.0110ME=?VCalculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.796 M and [Sn2 ] = 0.0170 M. Standard reduction potentials can be found here.Mg(s)+Sn2+(aq)\rightleftharpoonsMg2+(aq)+Sn(s)E=?V
Answer:
I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)
Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)
a) E(Pt⁺²/Fe°) = - 1.668v
b) Process is Non-spontaneous if E(cell) < 0
Explanation:
Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔
Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)
As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.
E°(Fe⁺²) = -0.44v
E°(Pt⁺²) = +1.20v
E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)
= -0.44v - (+1.20v) = - 1.64v
[Fe⁺²] = 0.0066M
[Pt⁺²] = 0.057M
n = electrons transferred = 2
E(nonstd) = E°(std) - (0.0592/n)logQ);
Q = [Pt⁺²]/[Fe⁺²]
= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v
Also, if ΔG(cell) > 0 => indicates non-spontaneous process
ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)
The Nernst equation is used to obtain the cell potential under standard conditions from the cell potential under nonstandard conditions.
Pt(s) + Fe^2+(aq) -------> Pt^2+ (aq) + Fe(s)
We have the following information from the question;
[Fe2+] = 0.0066M, [Pt2+] = 0.057M
The standard reaction potential is; 1.18 V - (-0.44V) = 1.62 V
Using Nernst equation;
Ecell = E°cell - 0.0592/n log Q
Where;
Q = [Pt2+] /[Fe2+] = 0.057M/ 0.0066M = 8.636
Ecell = 1.62 V - 0.0592/2 log(8.636)
Ecell = 1.59 V
Cu(s) + 2Ag+(aq) -------> Cu2+(aq) + 2Ag(s)
The standard reaction potential is; 0.80 V - 0.34 V = 0.46 V
Q = [Cu2+]/[Ag+]^2 = 0.013/0.013^2 = 76.92
Ecell = 0.46 V - 0.0592/2 log(76.92)
Ecell = 0.40 V
Co2+(aq) + Ti3+(aq) -------> Co3+(aq) + Ti2+(aq)
The standard reaction potential of the reaction is; 1.92 V - (-0.37) = 2.29 V
Q = [Ti2+] [Co3+]/[Co2+] [Ti3+] = [0.0110] [0.030]/[0.050] [0.0055] = 0.00033/0.000275 = 1.2
Ecell = 2.29 V - 0.0592/1 log(1.2)
Ecell = 2.28 V
Mg(s) + Sn2+(aq) --------> Mg2+(aq) + Sn(s)
The standard reaction potential of the reaction is; (-0.14 V) - (-2.37 V) = 2.23 V
Q = [Mg^2+]/[Sn^2+] = [0.796 M]/[0.0170 M] = 46.8
Ecell = 2.23 V - 0.0592/2 log(46.8)
Ecell = 2.18 V
Each of the cell is spontaneous as written since Ecell in each case is positive.
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What is the ratio of molecules in:
CH4 + 2O2 -> CO2 + 2H2O
Answer:
1:2:1:2
Explanation:
A molecule is made up of aggregates of small particles called atoms.
CH₄ + 2O₂ → CO₂ + 2H₂O
From the reaction above we have: 1 molecule of methane CH₄ reacting with 2 molecules of oxygen gas O₂ to produce 1 molecule of carbon dioxide gas CO₂ and 2 molecules of H₂O:
CH₄ + 2O₂ → CO₂ + 2H₂O
1 : 2 : 1 : 2
Hydrocarbons are organic compounds composed entirely of hydrogen and carbon. A 0.1647 g sample of a pure hydrocarbon was burned in a combustion apparatus to produce 0.4931 g of CO2 and 0.2691 g of water. Determine the empirical formula; enter as C#H# (for example: C1H1, write "1" if appropriate) Through another experiment it was determined that the molecular weight of this hydrocarbon is approximately 132 amu. What is the molecular formula of this compound? Enter as C#H# (for example C2H6, write "1" if appropriate)
The empirical formula of the hydrocarbon is C1H3, calculated from the products of combustion. Using the molecular weight approximation of 132 amu, the molecular formula determined is C8H24.
Explanation:To determine the empirical formula of the hydrocarbon, we analyze the products from its combustion. Since the sample produced 0.4931 g of CO2, we calculate the moles of carbon:
0.4931 g CO2 × (1 mol CO2/44.01 g CO2) = 0.01120 mol CO2
0.01120 mol CO2 corresponds to 0.01120 mol C (since there is 1 C in each CO2 molecule).
Similarly, given 0.2691 g of H2O:
0.2691 g H2O × (1 mol H2O/18.02 g H2O) = 0.01493 mol H2O
This gives us 0.02986 mol H (since there are 2 H in H2O).
The molar ratio of C to H can be simplified by dividing by the smaller number of moles:
Molar ratio C:H = 0.01120 mol C : 0.02986 mol H
We reduce this ratio to the simplest whole numbers to get the empirical formula:
(0.01120 / 0.01120) : (0.02986 / 0.01120) = 1:2.67
Approximating to whole numbers, we get C1H3 as the empirical formula.
To find the molecular formula, we use the given molecular weight (approximately 132 amu). Since the empirical formula weight of C1H3 is 12 (for C) + 3 (for H) = 15 amu, we calculate the multiplier:
Molecular weight / Empirical formula weight = 132 amu / 15 amu ≈ 8.8
This multiplier indicates the molecular formula is approximately 8 or 9 times the empirical formula. A whole number will result when using 8, thus we get the molecular formula C8H24.
Using the ideas from this section and the periodic table, choose the more reactive metal
K or Cu
Answer: K
Hope this helps you out!
Answer:
Potassium (K)
Explanation:
The reason that potassium is the more reactive metal is because it is in group 1, as opposed to group 11 for copper, and being in group 1 means that it has one valence electron, which it can give away by reacting with almost any other element, and in doing so, it becomes much more stable.
What are the correct half reactions for the following reaction:
Zn + 2 HCl -> H 2 + ZnCl 2
Answer:
D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.
Explanation:
It is a redox reaction that is consisted of two half-reactions:Oxidation reaction:
Zn losses 2 electrons and is oxidized to Zn²⁺:
Zn → Zn²⁺ + 2e⁻.
Reduction reaction:
H⁺ gains 1 electron and is reduced to H:
2H⁺ + 2e⁻ → H₂.
So, the right choice is: D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.
Answer:
D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.
Explanation:
Answer via Educere/ Founder's Education
Determine the pH of a HCl acid solution of 0.024
A. 0.62
B. 7.02
C. 1.62
D. 2.94
Answer:
C. 1.62.
Explanation:
∵ pH = - log[H⁺],
[H⁺] = 0.024 M.
∴ pH = - log(0.024) = 1.62.
So, the right choice is: C. 1.62.
For each pair below, select the sample that contains the largest number of moles. Pair A 2.50 g O2 2.50 g N2
Answer:
Explanation:
Pair 2.50g of O₂ and 2.50g of N₂
The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:
Number of moles = [tex]\frac{mass }{molarmass}[/tex]
Atomic mass of O = 16g and N = 14g
Molar mass of O₂ = 16 x 2 = 32gmol⁻¹
Molar mass of N₂ = 14 x 2 = 28gmol⁻¹
Number of moles of O₂ = [tex]\frac{2.5}{32}[/tex] = 0.078mole
Number of moles of N₂ = [tex]\frac{2.5}{28}[/tex] = 0.089mole
We see that N₂ has the largest number of moles
Answer:
Pair A : 2.50 g N2 Pair B : 21.5 g n2 Pair C : 0.081 CO2
Explanation:
Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is the natural abundance of Cl-35? (The atomic mass of chlorine is 35.45 amu)a) 75.95%b) 24.05%c) 50.00%d) 35.00%e) 37.00%
Final answer:
The natural abundance of Cl-35 is found by solving for the fraction x in the average atomic mass equation, which corresponds to the percentage. The solution shows that the natural abundance of Cl-35 is 75.77%.
Explanation:
The question is asking us to determine the natural abundance of Cl-35 given the average atomic mass of chlorine and the exact masses of its two stable isotopes, Cl-35 and Cl-37. To find this, we can use the formula for calculating the average atomic mass:
Average atomic mass = (fraction of isotope 1 × mass of isotope 1) + (fraction of isotope 2 × mass of isotope 2)
Let x be the fraction of Cl-35 and (1-x) the fraction of Cl-37. Since the percentages must add up to 100, we have:
35.45 amu = (x × 34.9689 amu) + [(1 - x) × 36.9695 amu]
Rearrange to solve for x:
x = (35.45 - 36.9695) / (34.9689 - 36.9695)
Calculate x and then convert x to a percentage:
x × 100 = percent abundance of Cl-35
Using x = 0.7577 (from given information), we find:
Percent abundance of Cl-35 = 0.7577 × 100 = 75.77%
Therefore, the natural abundance of Cl-35 is 75.77%.
Addition of _____________ to pure water causes the least increase in conductivity. A) weak bases B) acetic acid C) ionic compounds D) organic molecules
Organic molecules cause the least increase in conductivity when added to pure water, as they do not dissociate into ions, unlike ionic compounds which dissociate completely, and weak acids and bases which partially ionize.
The addition of organic molecules to pure water causes the least increase in conductivity. This is because organic molecules typically do not dissociate into ions when dissolved in water. On the other hand, ionic compounds dissociate almost completely in water, making them strong electrolytes and significantly increasing water's conductivity. Option D is correct .
Weak bases and acetic acid, a weak acid, only partially ionize in water, which would lead to a slight increase in conductivity compared to organic molecules, but much less than that caused by ionic compounds.
The weak bases mentioned do not refer to the dissociation of ionic compounds themselves but to the reactions of their polyatomic anions with water. Thus, when discussing weak bases in this context, it is important to consider these secondary reactions rather than primary dissociation.
Furthermore, acetic acid, being a weak acid, increases the hydronium ion concentration in water, but again not as much as a strong acid would. However, since both acetic acid and weak bases only partially ionize, they’re still considered weak electrolytes.
You want to clean a 500.-ml flask that has been used to store a 0.900M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.000100 M or below?
Answer:
At least four such rinses.
Explanation:
What's the concentration after each rinse?Let [tex]c_n[/tex] denotes the concentration of the residue after the [tex]n[/tex]th rinse. [tex]n[/tex] can be any non-negative integer (which includes zero.)
The initial concentration of the solution in the flask is [tex]\rm 0.900\;M[/tex]. That is:
[tex]c_0 = \rm 0.900\;M[/tex].
[tex]\rm 1.00\;mL[/tex] of this [tex]\rm 0.900\;M[/tex] solution is left in the flask after it is emptied for the first time.
The [tex]\rm 9.00\;mL[/tex] solvent will increase the volume of the solution from [tex]\rm 1.00\;mL[/tex] to [tex]\rm 10.00\;mL[/tex]. At this point the number of moles of solute in the flask stays unchanged. The concentration of the residue will drop to [tex]1/10[/tex] of the initial value. That is:
[tex]\displaystyle c_1 = \rm \frac{1}{10}\;c_0 = \frac{1}{10}\times 0.900\;M[/tex].
Repeat this process, and the concentration of the residue will drop by a factor of [tex]1/10[/tex] again. That is:
[tex]\displaystyle c_2 = \rm \frac{1}{10}\;c_1 = \frac{1}{10}\times (\frac{1}{10}\times 0.900\;M) = {\left(\frac{1}{10}\right)}^{2}\times 0.900\;M[/tex].
Summarize these values in a table:
[tex]\begin{array}{l|cccc}\text{Number of Rinses} & 0 & 1 & 2 & \dots \\[0.5em]\displaystyle\text{Concentration}\atop\displaystyle\text{of the residue}& \rm 0.900\;M & \rm\dfrac{1}{10}\times 0.900\;M &\rm {\left(\dfrac{1}{10}\right)}^{2}\times 0.900\;M \dots \end{array}[/tex].
The trend in [tex]c_{n}[/tex] is similar to that of a geometric series with
The concentration of the residue before any rinse, [tex]c_{0} = \rm 0.900\;M[/tex], andCommon ratio [tex]\displaystyle r = \frac{1}{10}[/tex].Again, refer to the trend in [tex]c_{n}[/tex] as the value of [tex]n[/tex] increases. The general formula for [tex]c_{n}[/tex], the concentration after the [tex]n[/tex]th rinse, will be:
[tex]\displaystyle c_{n} = \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M[/tex].
As a side note, [tex]0 < 1/10 < 1[/tex]. As a result, the value of [tex]c_{n}[/tex] will decrease but stay positive as the value of [tex]n[/tex] increases. Increasing the number of rinses will indeed reduce the concentration of the residue.
How many such rinses are required?In other words, what's the minimum value of [tex]n[/tex] for which [tex]\c_{n} \le \rm 0.000100\;M[/tex]?
Recall that
[tex]\displaystyle c_{n} = \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M[/tex].
As a result, [tex]n[/tex] should satisfy the condition:
[tex]\displaystyle \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M \le 0.000100\;M[/tex].
Multiply both sides by
[tex]\displaystyle \frac{1}{\rm 0.900\;M}[/tex],
which is positive and will not change the direction of the inequality:
[tex]\displaystyle \rm {\left(\frac{1}{10}\right)}^{n} \le \frac{0.000100}{0.900}[/tex].
Take the natural logarithm [tex]\ln[/tex] of both sides of the inequality. The function [tex]\ln(x)}[/tex] is increasing as [tex]x[/tex] increases on the range [tex]x > 0[/tex]. This function will not change the direction of the inequality, either.
[tex]\displaystyle \rm \ln\left[{\left(\frac{1}{10}\right)}^{n}\right] \le \ln\left(\frac{0.000100}{0.900}\right)[/tex].
Apply the power rule of logarithms: [tex]n[/tex] is an exponent inside the logarithm. That will be equivalent to an expression where [tex]n[/tex] is a coefficient in front of the logarithm operator:
[tex]\displaystyle \rm \ln\left[{\left(\frac{1}{10}\right)}^{n}\right] = n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right][/tex].
[tex]\displaystyle n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right]\le \ln\left(\frac{0.000100}{0.900}\right)[/tex].
Multiply both sides by
[tex]\displaystyle \frac{1}{\ln \left(\dfrac{1}{10}\right)}}[/tex].
Keep in mind that logarithms of numbers between 0 and 1 (excluding 0 and 1) are negative. The reciprocal of a negative number is still negative. The direction of the inequality will change. That is:
[tex]\displaystyle n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right]\times \frac{1}{\ln \left(\dfrac{1}{10}\right)}}\ge \ln\left(\frac{0.000100}{0.900}\right)\times \frac{1}{\ln \left(\dfrac{1}{10}\right)}}[/tex].
The right-hand side is approximately 3.95. However, [tex]n[/tex] has to be an integer. The smallest integer that is greater than 3.95 is 4. In other words, it takes at least four such rinses to reduce the residual concentration to under 0.000100 M (0.000090 M to be precise.) Three such rinses will give only 0.00900 M.
Final answer:
Through serial dilution principles, the minimum number of rinses needed to dilute the residual concentration of a solution in a flask to 0.000100M or below is approximately 7, considering a 1:10 dilution ratio for each rinse.
Explanation:
The question involves calculating the minimum number of rinse cycles required to reduce the residual concentration of a solution adhering to the walls of a flask below a target concentration, using serial dilution principles.
After each rinse, the concentration is diluted by the volume ratio, in this case, a 1:10 dilution per rinse because 1.00 ml of the original solution is mixed with 9.00 ml of solvent. The formula for the final concentration (Cf) after n rinses is given by Cf = Co × (Vo / Vt)n, where Co is the original concentration, Vo is the original volume adhering to the flask, Vt is the total volume after adding solvent, and n is the number of rinses.
To solve for n, we rearrange the equation to get n = log(Cf/Co) / log(Vo/Vt). Plugging in the given values (Co=0.900M, Cf=0.000100M, Vo=1ml, Vt=10ml), we find that the minimum number of rinses needed to reduce the concentration to 0.000100 M or below is calculated to be around 7. The exact number might slightly vary depending on rounding during calculation, but 7 rinses ensure meeting the target concentration threshold.
WILL MARK BRAINLIEST
If a planet has a lower orbital radius, what does that mean about its proximity to the sun? How would that affect the climate on that planet?
Answer:
If the planet has a lower orbital radius it means the planets proximity to the sun is farther away. The climate of the planet will be colder than a planet that would have a closer proximity to the sun.
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 17.0cm wide and 20.4cm high. The maximum safe pressure inside the vessel has been measured to be 2.20MPa. For a certain reaction the vessel may contain up to 0.0985kg of boron trifluoride gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Round your answer to significant digits.
Answer:
The maximum safe operating temperature the engineer should recommend for this reaction is 590. °C (3 significant figures)Explanation:
1) Data:
a) Cylinder diameter: d = 17.0 cm
b) Cylinder height: H = 20.4 cm
c) p = 2.20 MPa
d) compound: BF₃ gas
e) mass, m = 0.0985 kg
2) Formulae:
a) Volume of a cylinder, V = π r² H
b) Number of moles, n = mass in grams / molar mass
c) Ideal gas equation: pV = nRT
3) Solution:
a) Volume (V)
i) r = d / 2 = 17.0 cm / 2 = 8.50 cm
ii) V = V = π r² H = π (8.50 cm) ² (20.4 cm) = 4,630 cm³ = 4.630 liter
b) Number of moles (n)
i) molar mass BF₃: 67.82 g/mol
ii) n = mass in grams / molar mass = 98.5 g / 67.82 g/mol = 1.452 mol
c) Temperature
i) pressure conversion: 2.20 MPa = 22.21 atm
ii) pV = nRT ⇒ T = pV / (nR)
T = 22.21 atm × 4.630 liter / (1.452 mol × 0.08206 atm-liter /K-mol(T = 863 K T = 863 - 273 °C = 590. °C ← answerThe result is given with 3 significant figures, since that is the number of significant figures used for the data.
To calculate the maximum safe operating temperature for the gas reaction vessel, use the ideal gas law to find the number of moles of boron trifluoride gas and then calculate the maximum safe operating temperature.
Explanation:To calculate the maximum safe operating temperature for the gas reaction vessel, we need to consider the ideal gas law and the given quantities of gas. First, we convert the pressure from megapascals to pascals. Then, we use the ideal gas law equation to find the number of moles of boron trifluoride gas. Finally, we use the molar mass of boron trifluoride to convert the number of moles to grams, and then divide by the volume of the vessel to find the density. By rearranging the ideal gas law equation to solve for temperature, we can calculate the maximum safe operating temperature.
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Which value is MOST likely to be the boiling point of the substance at this constant pressure?
Answer:
48 c
Explanation:
its b on usa test prep
The thermostat in a refrigerator filled with cans of soft drinks malfunctions and the temperature of the refrigerator drops below zero celcius. The contents of the cans of diet soft drinks freeze, rupturing many of the cans and causing an awful mess. However, none of the cans containing regular non diet soft drinks rupture. Select the statement or statements below that best describe this behavior.A. As the temperature drops the solubility of the dissolved carbon dioxide gas decreases in the diet soft drinks. The pressure caused by this released gas builds up and finally ruptures the can.B.Water expands on freezing. Since water is the principle ingredient in soft drinks when the soft drink freezes it will rupture the can.C. There is more water in diet soft drink so they will freeze point of the solution sufficiently so that the solution does not freeze.D. Diet soft drink are inherently messier than non diet soft drinks.
The behavior of the cans of soft drinks in a malfunctioning refrigerator is explained by the properties of water's expansion on freezing and gas solubility changes under varying pressure and temperature, but none of the given options accurately explains why only diet soft drink cans ruptured.
Explanation:The scenario with cans of soft drinks in a malfunctioning refrigerator can be explained by the behavior of carbonated beverages when exposed to changes in temperature and pressure. The key statements that describe this behavior are:
Water expands on freezing. Since water is the principal ingredient in soft drinks, when the soft drink freezes, it will rupture the can. This is a general property of water and applies to all the cans equally. However, it does not directly explain why only the diet soft drinks' cans ruptured.Solution is exposed. Gas solubility in a liquid increases as the pressure of the gas over the liquid increases. During the carbonation process, beverages are exposed to high carbon dioxide pressure, saturating the beverage with CO₂. When the pressure is released, as in opening a can, a decrease in CO₂ solubility is observed, causing the release of gas. This principle also applies when the solubility of carbon dioxide decreases due to temperature drop, potentially causing cans to burst if the internal pressure becomes too high. However, this does not directly address the difference between diet and regular soft drinks.The actual reason for the difference in the behavior between diet and regular soft drinks is not explicitly stated in the options; however, the sugar content in regular soft drinks could play a role in depressing the freezing point of the solution, making it less likely to freeze and expand within the same temperature range.Therefore, while 'B. Water expands on freezing' is a correct statement, it does not fully explain why only diet soft drink cans ruptured. The differences in freezing point depression due to differing ingredient compositions between diet and regular soft drinks could provide a better explanation, albeit not directly addressed among the options given.
What kind of reaction occurs when a molecule of glucose reacts with oxygen to give carbon dioxide and water?
Answer:
An oxidation-reduction reaction
Explanation:
[tex]\stackrel{\hbox{-2}}{\hbox{C}}\text{$_{6}$H$_{12}$O$_{6}$} + 6\stackrel{\hbox{0}}{\hbox{O}\text{$_{2}$}} \longrightarrow 6\stackrel{\hbox{+4}}{\hbox{C}}\text{O$_{2}$}} + 6\text{H$_{2}$}\stackrel{\hbox{-2}}{\hbox{O}}[/tex]
Look at the oxidation numbers of the atoms.
Each carbon atom in glucose loses four electrons (oxidation), and each oxygen atom in O₂ gains two electrons (reduction).
Ba(OH)2 Ba+2 + 2 OH- (dissolved in solution). Which will NOT happen to the equilibrium of this solution as H+ ions are added? H+ will combine with OH- to form water. The base will dissociate to form more OH-. The reaction will move to the right. The reaction will move to the left. The quantity of Ba(OH)2 will decrease.
Answer:
The reaction will move to the left.
Explanation:
For the reaction:Ba(OH)₂ = Ba²⁺ + 2OH⁻,
Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.
If H⁺ ions are added to the equilibrium:H⁺ will combine with OH⁻ to form water.
So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.
According to Le Châtelier's principle: when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.
Answer:
D.) The reaction will move to the left.
Explanation: